/*
    Given int array & sliding window size k, return max sliding window
    Ex. nums = [1,3,-1,-3,5,3,6,7] k = 3 -> [3,3,5,5,6,7]

    Sliding window deque, ensure monotonic decr, leftmost largest

    Time: O(n)
    Space: O(n)
*/

class Solution {
public:
    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        deque<int> dq;
        vector<int> result;
        
        int i = 0;
        int j = 0;
        
        while (j < nums.size()) {
            while (!dq.empty() && nums[dq.back()] < nums[j]) {
                dq.pop_back();
            }
            dq.push_back(j);
            
            if (i > dq.front()) {
                dq.pop_front();
            }
            
            if (j + 1 >= k) {
                result.push_back(nums[dq.front()]);
                i++;
            }
            j++;
        }
        
        return result;
    }
};